Magnetism & Electromagnetism

Magnets form a magnetic field around them, caused by magnetic “poles.” These are similar to electric “poles” or “charge.”

Magnetic field lines leave the magnet from the north pole and reenter into the south pole

Magnetic field lines are continuous field lines in loops with no beginning or end (not like electric field lines

The symbol for a magnetic field is B

If they are allowed to select their own orientation, magnets align so that the north pole points in the direction of the magnetic field

Compasses are magnets that can easily rotate and align themselves

A compass points to the Earth’s North magnetic pole (which is near the north geographic pole)

Is the North Magnetic Pole the north pole of the Earth’s Magnetic field?

Magnetic Monopoles DO NOT EXIST!!!• Magnetic poles cannot be separated from

each other in the same way that electric poles (charges) can be

• Electric monopoles exist as either a negatively charged object or a positively charged object

Units:•Tesla (SI unit)N/(C m/s)N/(A m)

•Gauss1 Tesla = 104 gauss

Magnetic fields cause the existence of magnetic forces like electric fields cause electric forces

A magnetic force is exerted on a particle within a magnetic field only if• The particle has a charge• The charged particle is moving with at least

a portion of its velocity PERPENDICULAR to the magnetic field

Magnitude: F = qvBsinΘ• q = charge in Coulombs (C)• v = velocity in m/s• B = magnetic field in Tesla• Θ = angle between v and B

Direction: Right hand rule if q is positive, left hand rule if q is negative

FB = q v x B (This is a “vector cross product” for those of you who know your math)

Direction of the magnetic force?

Right Hand RuleTo determine the DIRECTION of the

force on a POSITIVE charge we use a special technique that helps us understand the 3D/perpendicular nature of magnetic fields.

Basically you hold your right hand flat with your thumb perpendicular to the rest of your fingers

•The Fingers = Direction B-Field•The Thumb = Direction of velocity•The Palm = Direction of the Force

For NEGATIVE charges use left hand!

Calculate the magnitude of the force on a 3.0 C charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed• North• East • South• West

• Sample Problem:– Calculate the magnitude of force exerted on a 3.0 μC

charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed a)N, b)E, c)S, d)W

a) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin0m/s ∙ 0.2T ∙ sin0oo = 0 N= 0 N

b) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin90m/s ∙ 0.2T ∙ sin90oo = 0.18 N= 0.18 Nc) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin180m/s ∙ 0.2T ∙ sin180oo = 0 N= 0 Nd) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin270m/s ∙ 0.2T ∙ sin270oo = -0.18 N= -0.18 N

Calculate the magnitude and direction of the magnetic force

• Calculate the magnitude and direction of the magnetic force.

v = 300,000 m/s

B = 200 mTq = 3.0μC

34o

F = qvBsin= qvBsinθθF = 3x10= 3x10-6-6C ∙ 300000 m/s ∙ 0.2T ∙ sin(34C ∙ 300000 m/s ∙ 0.2T ∙ sin(34oo) )

F =0.101N upward

Magnetic forces are always orthogonal (at right angles) to the plane established by the velocity and magnetic field vectors

Magnetic forces can accelerate charged particles by changing their direction

Magnetic forces can cause charged particles to move in circular or helical paths

Magnetic Forces CANNOT change the speed or KE of charged particles

Magnetic Forces CANNOT do work on charged particles (F is perpendicular)

Magnetic forces ARE centripetal•Remember centripetal

acceleration is v2/r•Centripetal force is mv2/r

F = maFB = Fc

qvBsin = mv2/rqB = mv/r

q/m = v/(rB)

B

F

V

F

V

F

V

F

V

What is the orbital radius of a proton moving at 20,000 m/s perpendicular to a 40 T magnetic field?

Magnetic Forces on Charged Particles …

• …are centripetal.• Remember centripetal force is mv2/r.• For a charged particle moving perpendicular

to a magnetic field– F = qvB = mv2/r

• Radius of curvature of the particle– r = mv2/qvB = mv/qB

Sample ProblemWhat is the orbital radius of a proton moving at 20,000 m/s perpendicular to a 40 T magnetic field?

What must be the speed of an electron if it is to have the same orbital radius as the proton in the magnetic field described in the previous problem?

Sample ProblemWhat must be the speed of an electron if it is to have the same orbital radius as the proton in the magnetic field described in the previous problem?

An electric field of 2,000 N/C is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton. Ignore gravitational effects.

Sample ProblemSample ProblemAn electric field of 2000 N/C is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton? Ignore gravitational effects.

A magnetic field of 2,000 mT is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton. Ignore gravitational effects.

Sample ProblemSample ProblemA magnetic field of 2000 mT is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton? Ignore gravitational effects.

Calculate the force and describe the path of this electron if the electric field strength is 2000 N/C

Sample ProblemSample Problem

Calculate the force and describe the path of this electron.

E = 2000 N/C

e-300,000 m/s

How would you arrange a magnetic field and an electric field so that a charged particle of velocity v would pass straight through without deflection?

Electric and Magnetic Fields Electric and Magnetic Fields TogetherTogether

E

B

v = E/B

e-

It is found that protons when traveling at 20,000 m/s pass undeflected through the velocity filter below. What is the magnetic field between the plates?

0.02 m

400 V

e+

Sample ProblemSample ProblemIt is found that protons traveling at 20,000 m/s pass undeflected through the velocity filter below. What is the magnitude and direction of the magnetic field between the plates?

400 V

e 20,000 m/s0.02 m

F = I L B sin Θ• I = current in Amps• L = length in m• B = magnetic field in Tesla• Θ = angle between current and B field

What is the force on a 100m long wire bearing a 30A current flowing north if the wire is in a downward-directed magnetic field of 400 mT?

Sample ProblemSample ProblemWhat is the force on a 100 m long wire bearing a 30 A current flowing north if the wire is in a downward-directed magnetic field of 400 mT?

What is the magnetic field strength if the current in the wire is 15 A and the force is downward with a magnitude of 40 N/m? What is the direction of the current?

Sample ProblemSample ProblemWhat is the magnetic field strength if the current in the wire is 15 A and the force is downward and has a magnitude of 40 N/m? What is the direction of the current?

Magnetic Fields affect moving charge• F = qvBsinΘ• F = ILBsin Θ

Magnetic fields are caused by moving charge

B = 0I/(2πr)0 : 4 π x 10-7 T m/A Magnetic permeability of free space

• I: current (A)•R: radial distance from center of wire (m)

1. Curve your fingers2. Place your thumb in the

direction of the current3. Curved fingers

represent the curve of the magnetic field

4. Field vector at any point is tangent to the field line

What is the magnitude and direction of the magnetic field at P, which is 3.0 m away from a wire bearing a 13.0 A current?

Sample ProblemSample ProblemWhat is the magnitude and direction of the

magnetic field at point P, which is 3.0 m away from a wire bearing a 13.0 Amp current?

I = 13.0 A

P3.0 m

What is the magnitude and direction of the force exerted on a 100 m long wire that passes through point P which bears a current of 50 Amps in the same direction?

P

I1 = 13.0 A

I2 = 50.0 A

3.0 m

Sample Problem – not in packetSample Problem – not in packetWhat is the magnitude and direction of the

force exerted on a 100 m long wire that passes through point P which bears a current of 50 amps in the same direction?

I1 = 13.0 A

P

3.0 m

I2 = 50.0 A

Remember this from electrostatics? When there are two or more currents

forming a magnetic field, calculate B due to each current separately, and then add them together using vector addition.

16. What is the magnitude and direction of the electric field at point P if there are two wires producing a magnetic field at this point?

P

I1 = 13.0 A

I2 = 10.0 A

3.0 m

4.0 m

Sample ProblemSample ProblemWhat is the magnitude and direction of the

electric field at point P if there are two wires producing a magnetic field at this point?

I = 13.0 A

P3.0 m

I = 10.0 A4.0 m

You learned that coils with current in them make magnetic fields (electromagnets)

The iron nail was not necessary to cause the field, it only intensified it

A solenoid is a coil of wire When current runs through a wire, it

causes the coil to become an “electromagnet” Air-core solenoids

have nothing inside fo them

Iron-core solenoids are filled with iron (a magnetic material) to intensify the magnetic field

B

1. Curve your fingers2. Place them along the wire

loop so that your fingers point in the direction of the current

3. Your thumb gives the direction of the magnetic field in the center of the loop, where it is straight

4. Field lines curve around and make complete loops

What is the direction of the magnetic field produced by the current I at A? At B?

Sample ProblemSample ProblemWhat is the direction of the magnetic field produced by the current I at A? At B?

I

AB

The product of magnetic field and area

Can be thought of as total magnetic “effect” on a coil of wire of a given area

The area is aligned so that a perpendicular to the area vector (orthogonal to area) points parallel to the field

The area is aligned so that a perpendicular to the area vector points perpendicular to the field

The area is neither perpendicular nor parallel

ΦB = B A cos Θ• ΦB : magnetic flux in Webers (Tesla meters2)• B: magnetic field in Tesla• A: area in meters2

• Θ : the angle between the area vector and the magnetic field

ΦB = B A

Calculate the magnetic flux through a rectangular wire frame 3.0 m long and 2.0 m wide if the magnetic field through the frame is 4.2 mT• Assume that the magnetic field is

perpendicular to the area vector• Assume that the magnetic field is parallel to

the area vector• Assume that the angle between the

magnetic field and the area vector is 30 degrees

Sample ProblemSample Problem

Calculate the magnetic flux through a rectangular wire frame 3.0 m long and 2.0 m wide if the magnetic field through the frame is 4.2 mT.

a) Assume that the magnetic field is perpendicular to the area vector.

b) Assume that the magnetic field is parallel to the area vector.

c) Assume that the angle between the magnetic field and the area vector is 30o.

Assume the angle is 40 degrees, the magnetic field is 50 mT, and the flux is 250 mWb. What is the radius of the loop? (hint: A = πr2)

Sample ProblemSample ProblemAssume the angle is 40o, the magnetic field is 50

mT, and the flux is 250 mWb. What is the radius of the loop?

BA

A system will respond so as to oppose changes in magnetic flux

A change in magnetic flux will be partially offset by an induced magnetic field whenever possible

Changing the magnetic flux through a wire loop causes current to flow in the loop

This is because changing magnetic flux induces an electric potential

ε = -NΔΦB/Δt• ε : induced potential (V)• N : # loops• ΦB : magnetic flux in Wb• t : time (s)

If there is only ONE LOOP: ε = -ΔΦB/Δt ε = -Δ(B A cos Θ)/Δt

•To generate voltage: Change B Change A Change Θ

A coil of radius 0.5 m consisting of 1000 loops is placed in a 500 mT magnetic field such that the flux is maximum. The field then drops to zero in 10 ms. What is the induced potential in the coil?

Sample ProblemSample ProblemA coil of radius 0.5 m consisting of 1000 loops is

placed in a 500 mT magnetic field such that the flux is maximum. The field then drops to zero in 10 ms. What is the induced potential in the coil?

A single coil of radius 0.25 m is in a 100 mT magnetic field such that the flux is maximum. At time t = 1.0 s, the field increases at a uniform rate so that at 11 s, it has a value of 600 mT. At time t = 11 s, the field stops increasing. What is the induced potential:a. At 0.5 s?b. At 3.0 s?c. At 12 s?

Sample ProblemSample ProblemA single coil of radius 0.25 m is in a 100 mT

magnetic field such that the flux is maximum. At time t = 1.0 seconds, field increases at a uniform rate so that at 11 seconds, it has a value of 600 mT. At time t = 11 seconds, the field stops increasing. What is the induced potential

A) at t = 0.5 seconds?B) at t = 3.0 seconds?C) at t = 12 seconds?

The current will flow in a direction so as to oppose the change in flux

Use in combination with the hand rule to predict current direction

The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction fo the current in the wire loop?

Sample ProblemSample ProblemThe magnetic field is increasing at a

rate of 4.0 mT/s. What is the direction of the current in the wire loop?

The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop?

Sample ProblemSample ProblemThe magnetic field is increasing at a

rate of 4.0 mT/s. What is the direction of the current in the wire loop?

The magnetic field is decreasing at a rate of 4.0 mT/s. The radius of the loop is 3.0 m, and the resistance is 4 ohms. What is the magnitude and direction of the current?

Sample ProblemSample ProblemThe magnetic field is decreasing at a rate

of 4.0 mT/s. The radius of the loop is 3.0 m, and the resistance is 4 . What is the magnitude and direction of the current?

ε = BLv• B: magnetic field• L: length of bar moving through field• v: speed of bar moving through field

Bar must be “cutting through” the field lines. It cannot be moving parallel to the field

This formula is derivable from Faraday’s Law of Induction

How much current flows through the resistor? How much power is dissipated by the resistor?

Sample ProblemSample ProblemHow much current flows through the

resistor? How much power is dissipated by the resistor?

B = 0.15 T

v = 2 m/s

50 cm3

In which direction is the induced current through the resistor?

Sample ProblemSample ProblemIn which direction is the induced current

through the resistor (up or down)?

B = 0.15 T

v = 2 m/s

50 cm3

Assume the rod is being pulled so that it is traveling at a constant 2 m/s. How much force must be applied to keep it moving at this constant speed?

Sample ProblemSample ProblemAssume the rod is being pulled so that it is

traveling at a constant 2 m/s. How much force must be applied to keep it moving at this constant speed?

B = 0.15 T

v = 2 m/s

50 cm3